2. Inequalities Solutions 2023 (RVHS)

River Valley High School, Mathematics Department 2023 Inequalities T1 Inequalities 1 2 4 132 x xx   ----------- (*) 2 4 1032 x xx   2 22 4 3 2 03 2 3 2 x x x x x x x       2 2 1 032 xx xx   2 2 1 023 xx xx   Since 2 2 1310 24x x x       for any real x (OR coefficient of 2 10x  and discriminant, 22 4 ( 1) 4(1)(1) 3 0b ac      imply that 2 10xx   for any real x) 2 1 023xx  1 0( 1)( 3)xx  1 or 3xx   By replacing x with   2x in (*), we have: 2 24 4 132 x xx   2 42 4 123 x xx   which is what we need to solve. From earlier part, 22 1 or 3xx    2 1x  1 or 1xx  – 1 3 x + + – (N.A., since 2 0x for all real x)

River Valley High School, Mathematics Department 2023 Inequalities T2 2i   224 4 3 2 1 2 0x x x x        Alternative Method Discriminant =    24 4 4 3 32  Since coefficient of 2x is positive and discriminant < 0, 24 4 3xx is positive ii Therefore or iii Let lnyx 32 2 4 4 3 02 y y y yy   20 y   or 1y 2 ln 0 x   or ln 1x  1x or ln 1x or ln 1x 11 or 0 e or ex x x     3 2 25 03 w w   OR 33 w   or 5 2w   2 2 5 sin 03 yx y   , Since sin 0x for 3 2x   , 2 25 03 y y   From above, 33 y   or 5 2y  03 y or 5 2y            32 2 2 2 4 4 3 02 4 4 3 021 0 since 2 1 2 021 x x x xx x x x xx x xxxx           2x 01 x 0 1 + − + -2 − – + – + 3 3 5 2 3 3 5 2

River Valley High School, Mathematics Department 2023 Inequalities T3 33 y   or 5 2y or 5 2y 4i         2 2 44 2 4 402 4 ( 2) 4 02 24 02 ( 1) 5 0( 2) 11 02 x x xx xx x xx x x x xx x                     2 or 1 1xx        Alternative       2 2 2 2 44 2 Multiply by ( 2) on both sides, (4 )( 2) 4( 2) 0, 2 ( 2)[(4 )( 2) 4] 0 ( 2)( 2 4) 0 ( 2)( 2 4) 0 ( 2) 1 1 0 x x x x x x x x x x x x x x x x x x x                                2 or 1 1xx        ii 44 2x x  – + – + 22 ( )( )a b a b a b    2x

River Valley High School, Mathematics Department 2023 Inequalities T4       441 12 45 1 Hence, replace with 1 in earlier sol: 1 2 or 1 1 1 1 or 2 2 as 0 2 and 2 for all real as 0 and (2 ) x x x x xx xx xx x x x x x x                                       No solution 2 i.e. 2 2 x x          5   2 2 3 2 6 11 3 2( 6 11) 11 202 x xx x x x xx          112 2x 11 11Let 2 ,2 2 0 ln 24 xxy e e x      O y x y = |x| y = −1 Since 2 5 0, |x| is always greater than 25 for all real x

River Valley High School, Mathematics Department 2023 Inequalities T5 6 ln 2 0 ax x   where 1a . 2 21 2 0 1 1 8 ) 1 1 8 )2 44 0 1 1 8 1 1 8 044 ax x x x a x aaxx x aa x or x                       7       2 2 2 2 2 23 0( 1) 1 1 1 3 0( 1)( 1) 12 0( 1)( 1) Since 1 2 0 for all , ( 1)( 1) 0 xx ax a x x ax x x ax x xx ax x                   If a is positive i.e. 01 a 11 x a If a is negative i.e. 0a 1 or 1.xx a 1 1 a 1 1 a

River Valley High School, Mathematics Department 2023 Inequalities T6 8       2 0x a x b x x c    , xc , 0x Since   2 0xc as xc ,    0x a x b x      0x x a x b   xa or 0 xb and xc Replace x by ln x . ln xa or 0 ln xb and ln xc 0 axe  or 1 bxe and cxe 9   2 2 2 0 1 xx x a x a      217() 24 0( )( 1) x x a x   Since numerator is always positive, 1 0 ( )( 1) 0 1( )( 1) x a x x ax a x         10              2 2 2 1 11 1 1 1 0 11 0 11 0 11 0 0 or 1 xxx xxx x x x x xx x xx x xx                 1 0 1    

River Valley High School, Mathematics Department 2023 Inequalities T7   21 e e e 0 11 e 1 ee Replace by e , e 1 or e 0 (reject) 0 x x x xx x x xx x x          11 3 541 xx       2 2 3 5 4 01 3 5 9 4 01 4 9 2 01 4 1 2 01 xx xx x xx x xx x            1 or 1 24xx    Let sinxy , 1sin or 1 sin 2 (rejected)4 0.253 or 2.89 yy yy           2 1 2 + + – –

River Valley High School, Mathematics Department 2023 Inequalities T8 12 Method 1: Simplifying the modulus using the graph From the graph, the root of  1 b x aax  is equal to the root of  1 b x aax       21 xab   1 .xa b    Since ,xa 1 .xa b  -------------------------------------------------------------------------- Method 2: Solve by squaring both sides  1 b x aax    2 221 b x aax             2 2 2 4 2 1 1 (since > 0) 1 . x a a x x ab x a bb xa b             Since ,xa 1 .xa b  From the sketch in part (i), xa or 1a x a b    OR 1 , x a x a b    x y O

River Valley High School, Mathematics Department 2023 Inequalities T9 13 2 3 1 2 1 3 02 1 1 3 02 1 1 ( 1) ( 3)(2 1) 0(2 1)( 1) ( 1) 0(2 1)( 1) xx xx xx xx xx xx x x x x xx x xx                1 2 1, 1xx    Replace x by x 1 2 1 2 11 22 1, 1 1 , 1 ,1 xx xx xx              14 22 22 ln( 4) 8 0 ln( 4) 8 xx xx        The suitable additional curve is 28yx . Intersection points between the 2 graphs: y 2x 2x   2ln 4yx 28yx x

River Valley High School, Mathematics Department 2023 Inequalities T10    2.63,1.07 , 2.63,1.07 (3 s.f.) 22The solution to the inequality ln( 4) 8 0 : 2.63 2 or 2 2.63 (3 s.f.) xx xx          Qn 15 TJC Prelim 9758/2022/01/Q3 (i) At the intersection points, 1x a ax    1 or 1 1 or 1 x a ax x a ax xx         From the graph, the solution is 11 x   (ii) 11Replacing by , 1 1x xx    From the graph of 1 ,y x The solution is 1 or 1xx  y x a 1 a y x y = 1 y = 1 1 1